poj2352，poj 1426 Find The Multiple （簡單搜索dfs）

題目：

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

Input

The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

poj2352，Output

For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

Sample Input

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

poj1426。?

題意：

????????本題要找出數m，m是只有0和1構成的十進制數，并且是n的倍數，若有多個答案，輸出任意一個就可以。

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題解：

??????? 經過思考會發現m最大不會超過unsigned long long 的范圍，所以用unsigned long long保存就可以，接下來就是深搜就ok。

dfs深度優先搜索。?

代碼：

#include <iostream>using namespace std;
unsigned long long ans;
bool f;void dfs(unsigned long long s,int n,int k)
{if(f) return ;if(s%n==0) {ans=s;f=true; return ;}if(k==19) return ;        //如果k超過19就不在unsigned long long的范圍內了dfs(s*10,n,k+1);dfs(s*10+1,n,k+1);return ;
}int main()
{int n;while(cin>>n,n){ans=0;f=false;dfs(1,n,0);cout<<ans<<endl;}return 0;
}

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